Question

Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800˚C (1073 K) is 3.6 × 1023 m-3 . The atomic weight and density (at 800˚C) for silver are, respectively, 107.9 g/mol and 9.5 g/cm3 .

Answer 1

Answer : The energy for vacancy formation in silver is, $1.762\times 10^{-19}J$

Explanation :

Formula used :

$N_v=N\times e^{(\frac{-E}{K\times T})}$

or,

$N=\frac{N_A\times \rho}{M}$

So,

$N_v=[\frac{N_A\times \rho}{M}]\times e^{(\frac{-E}{K\times T})}$

where,

$N_v$ = equilibrium number of vacancies = $3.6\times 10^{23}m^{-3}=3.6\times 10^{20}L^{-1}$

E = energy = ?

M = atomic weight = 107.9 g/mole

$N_A$ = Avogadro's number = $6.022\times 10^{23}mol^{-1}$

$\rho$ = density = $9.5g/cm^3=9.5g/ml=9500g/L$

T = temperature = $800^oC=273+800=1073K$

K = Boltzmann constant = $1.38\times 10^{-23}J/K$

Now put all the given values in the above formula, we get:

$3.6\times 10^{20}L^{-1}=[\frac{(6.022\times 10^{23}mol^{-1})\times 9500g/L}{107.9g/mol}]\times e^{[\frac{-E}{(1.38\times 10^{-23}J/K)\times 1073K}]}$

$E=1.762\times 10^{-19}J$

Therefore, the energy for vacancy formation in silver is, $1.762\times 10^{-19}J$

Answer 2

Answer:

The energy for vacancy formation in silver is 1.1 ev/atom

Explanation:

The total number of sites is equal to:

$N=\frac{N_{A} \rho }{A}$

Where

NA = Avogadro´s number = 6.023x10²³atom/mol

A = atomic weight of silver = 107.9 g/mol

ρ = density of silver = 9.5 g/cm³

Replacing:

$N=\frac{6.023x10^{23}*9.5 }{107.9} =5.3x10^{22} atom/cm^{3} =5.3x10^{28} m^{-3}$

The energy for vacancy is equal:

$Q=-RTln(\frac{N_{v} }{N} )$

Where

R = 8.314 J/mol K = 8.614x10⁻⁵ev/atom K

T = 800°C = 1073 K

Nv = number of vacancy = 3.6x10²³m⁻³

Replacing:

$Q=-8.614x10^{-5} *1073*ln(\frac{3.6x10^{23} }{5.3x10^{28} } )=1.1ev/atom$